Problem: Rewrite the function by completing the square. $f(x)=x^{2}+2x-39$ $f(x)=(x+$
Solution: We want to complete $x^2{+2}x$ into a perfect square. To do that, we should add $\left(\dfrac{{+2}}{2}\right)^2={1}$ to it: $x^2{+2}x+{1}=(x+1)^2$ In order to keep the expression equivalent, we add and subtract ${1}$, not forgetting the expression's constant term, $-39$ : $\begin{aligned} f(x)&=x^2+2x-39 \\\\ &=x^2+2x+{1}-39-{1} \\\\ &=(x+1)^2-39-1 \\\\ &=(x+1)^2-40 \end{aligned}$ In conclusion, after completing the square, the function is written as $f(x)=(x + 1)^2 - 40$